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mars1129 [50]
2 years ago
13

Identify the proper form of the equilibrium-constant expression for the equation

Chemistry
1 answer:
GaryK [48]2 years ago
6 0

Pure solids, liquids and gases are excluded from the equilibrium expression.

<h3>What is equilibrium constant?</h3>

The term equilibrium constant has to do with the number that shows the extent to which reactants are being converted into products.

In the first reaction; NO2(g)+NO3(g)⇌N2O5(g), the low value of the equilibrium constant implies that the reaction tends towards the left hand side.

In the second reaction, the equilibrium expression is [HOCl]^2. Note that pure solids, liquids and gases are excluded from the equilibrium expression.

Learn more about equilibrium constant:brainly.com/question/10038290

#SPJ1

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The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?
juin [17]

The formula for mole fraction is:

mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}    -(1)

The solubility of oxygen gas = 1.0 mmol/L  (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

Density = \frac{mass}{volume}

Mass = 1.0 g/ml\times 1000 mL = 1000 g

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = \frac{1000 g}{18 g/mol} = 55.55 mol

Substituting the values in formula (1):

mole fraction = \frac{0.001}{55.55+0.001}

mole fraction = 1.8\times 10^{-5}

Hence, the mole fraction is 1.8\times 10^{-5}.

7 0
3 years ago
Ann and Jesse's mom is driving them to school. They are listening to music on an FM radio station. When they ride in the car wit
Svet_ta [14]

Answer: it they are both in the same place

Explanation:I don’t know and don’t care loser

3 0
3 years ago
How you could adjust the settings of the simulation to increase the number of red (H3O+) and blue (A-) particles in the solution
Slav-nsk [51]

Answer:

this is good update to the doctor and I will be taking the test soon and will have a gf with my

8 0
3 years ago
Help please! What is the mass of the iron (Fe) involved in this reaction? Identify the following equations as balanced or unbala
eimsori [14]
<h2>Question 1:</h2><h3>32g + <u><em>56g</em></u> --> 88g</h3>

This one's just a simple math problem. Do 88 - 32, which is 56. You can also find the same answer by looking for the atomic mass of iron (Fe), which rounds up to 56.


<h2>Question 2:</h2><h3>This equation is <u><em>balanced</em></u>.</h3>

On the left side, there is 2 Nitrogen (because of the subscript), and 6 Hydrogen (3 x 2 = 6). On the right side, there is 2 Nitrogen (because of the coefficient), and 6 Hydrogen. (2 x 3 again) There are 2 Nitrogen and 6 Hydrogen on each side, therefore the equation is balanced.


<h2>Question 3:</h2><h3>This equation is <em><u>unbalanced.</u></em></h3>

On the left side, there are 2 Nitrogen (N) and 2 Hydrogen (H). On the right side, there is 1 Nitrogen and 6 Hydrogen. There are 2 Nitrogen and 2 Hydrogen on one side, while the other only has 1 Nitrogen and 6 Hydrogen. Therefore the equation is unbalanced.


<h2>Question 4:</h2><h3>This equation is <em><u>balanced.</u></em></h3>

On the left side, there are 2 Chromium (Cr) and 6 Chlorine (Cl). On the right side, there are 2 Chromium and 6 Chlorine. There are 2 Chromium and 6 Chlorine on each side, therefore the equation in balanced.

~~~~~~~~~~~~~~~~~~~~~~~

If something is confusing or doesn't make any sense, feel free to let me know! ^^

6 0
4 years ago
A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)

<u>Oxidation half reaction:</u> Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V       ( × 2)

<u>Reduction half reaction:</u> Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V       ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.16-(-1.66)=1.82V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = 42^oC=[42+273]K=315K

F = Faraday's constant = 96500

[Al^{3+}]=1.63M

[Cu^{2+}]=3.43M

Putting values in above equation, we get:

E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V

Hence, the cell voltage of the given reaction is 1.86 V

4 0
3 years ago
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